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            <h1 id="seo-header">『算法-ACM竞赛-图论』最大团问题</h1>
            
            
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                <h1 id="『算法-ACM-竞赛-图论』最大团问题"><a href="#『算法-ACM-竞赛-图论』最大团问题" class="headerlink" title="『算法-ACM 竞赛-图论』最大团问题"></a>『算法-ACM 竞赛-图论』最大团问题</h1><p><strong>一、定义</strong></p>
<p>一个无向图 G&#x3D;(V，E)，V 是点集，E 是边集。取 V 的一个子集 U，若对于 U 中任意两个点 u 和 v，有边 (u,v)∈E，那么称 U 是 G 的一个完全子图。 U 是一个团当且仅当 U 不被包含在一个更大的完全子图中。</p>
<p>G 的最大团指的是定点数最多的一个团。</p>
<p>简单来说，极大团是增加任一顶点都不再符合定义的团，最大团是图中含顶点数最多的极大团，最大独立集是除去图中的团后的点集，而最大团问题就是在一个无向图中找出一个点数最多的完全图。</p>
<p>二、常用结论</p>
<p>1、最大团点的数量&#x3D;补图中最大独立集点的数量</p>
<p>2、二分图中，最大独立集点的数量+最小覆盖点的数量&#x3D;整个图点的数量</p>
<p>3、二分图中，最小覆盖点的数量&#x3D;最大匹配的数量</p>
<p>4、图的染色问题中，最少需要的颜色的数量&#x3D;最大团点的数量</p>
<p><strong>三、算法实现</strong></p>
<p>毕竟是 NP 完全问题，所以具体使用，什么算法，区别不是很大，具体体现在剪枝上！</p>
<p>对于弦图来说，求最大团一般使用 MCS 算法，而对于一般图来说，常使用 Bron-Kerbosch 算法</p>
<p><strong>【Bron-Kerbosch 算法】</strong><br>Bron-Kerbosch 算法用于计算图中的最大的全连通分量，即计算图的最大团。</p>
<p><strong>1.算法原理</strong><br>Bron-Kerbosch 算法的基础形式是一个递归回溯的搜索算法，其通过给定三个集合：R、P、X 来递归的进行搜索</p>
<p>初始化集合 R、X 分别为空，集合 P 为所有顶点的集合<br>每次从集合 P 中取顶点 {vi}，当集合中没有顶点时，有两种情况：<br>1）集合 R 是最大团，此时集合 X 为空<br>2）无最大团，此时回溯<br>对于每一个从集合 P 中取得的顶点 {vi}，有如下处理：<br>1）将顶点 {vi} 加到集合 R 中，集合 P、X 与顶点 {vi} 得邻接顶点集合 N{vi} 相交，之后递归集合 R、P、X<br>2）从集合 P 中删除顶点 {vi}，并将顶点 {vi} 添加到集合 X 中<br>3）若集合 P、X 都为空，则集合 R 即为最大团<br>总的来看，就是每次从集合 P 中取 vi 后，再从 P∩N{vi} 集合中取相邻结点，保证集合 R 中任意顶点间都两两相邻</p>
<p>伪代码过程</p>
<figure class="highlight tp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br></pre></td><td class="code"><pre><code class="hljs tp">BronKerbosch<span class="hljs-number">1</span>(<span class="hljs-keyword">R</span>,<span class="hljs-keyword">P</span>,<span class="hljs-keyword">X</span>):<br>    if <span class="hljs-keyword">P</span> and <span class="hljs-keyword">X</span> are both empty:<br>        report <span class="hljs-keyword">R</span> as a maximal clique<br>    for each vertex v in <span class="hljs-keyword">P</span>:<br>        BronKerbosch<span class="hljs-number">1</span>(<span class="hljs-keyword">R</span> ⋃ &#123;v&#125;, <span class="hljs-keyword">P</span> ⋂ N(v), <span class="hljs-keyword">X</span> ⋂ N(v))<br>        <span class="hljs-keyword">P</span> := <span class="hljs-keyword">P</span> \ &#123;v&#125;<br>        <span class="hljs-keyword">X</span> := <span class="hljs-keyword">X</span> ⋃ &#123;v&#125;<br></code></pre></td></tr></table></figure>

<p><strong>2.算法优化</strong><br>对于基础的算法，由于其递归搜索了所有情况，对其中有些不是最大团的也进行了搜索，效率不高，为了节省时间让算法更快的回溯，可以通过设定关键点来进行搜索。</p>
<p>由于对于任意的最大团，其必须包括顶点 {u} 或 N-N{u}，不然其必然需要通过添加它们来进行扩充，这显然矛盾，所以仅需测试顶点 {u} 以及 N-N{u} 即可。</p>
<p>伪代码过程：</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br></pre></td><td class="code"><pre><code class="hljs cpp"><span class="hljs-built_in">BronKerbosch2</span>(R,P,X):<br>    <span class="hljs-keyword">if</span> P <span class="hljs-keyword">and</span> X are both empty:<br>        report R as a maximal clique<br>    choose a pivot vertex u in P ⋃ X<br>    <span class="hljs-keyword">for</span> each vertex v in P \ <span class="hljs-built_in">N</span>(u):<br>       <span class="hljs-built_in">BronKerbosch2</span>(R ⋃ &#123;v&#125;, P ⋂ <span class="hljs-built_in">N</span>(v), X ⋂ <span class="hljs-built_in">N</span>(v))<br>       P := P \ &#123;v&#125;<br>       X := X ⋃ &#123;v&#125;<br></code></pre></td></tr></table></figure>

<p>由于其是通过选择特殊点，来进行最小化递归调用，一定程度上节省了时间，但还可以与降序的方式结合使用，来保证在线性的时间内求子图的最大团</p>
<p><strong>伪代码过程：</strong></p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br></pre></td><td class="code"><pre><code class="hljs cpp"><span class="hljs-built_in">BronKerbosch3</span>(G):<br>    P = <span class="hljs-built_in">V</span>(G)<br>    R = X = empty<br>    <span class="hljs-keyword">for</span> each vertex v in a degeneracy ordering of G:<br>        <span class="hljs-built_in">BronKerbosch2</span>(R ⋃ &#123;v&#125;, P ⋂ <span class="hljs-built_in">N</span>(v), X ⋂ <span class="hljs-built_in">N</span>(v))<br>        P := P \ &#123;v&#125;<br>        X := X ⋃ &#123;v&#125;<br></code></pre></td></tr></table></figure>

<p><strong>3.实现</strong></p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br><span class="line">72</span><br><span class="line">73</span><br><span class="line">74</span><br></pre></td><td class="code"><pre><code class="hljs cpp"><span class="hljs-type">int</span> n,m;<br><span class="hljs-type">bool</span> G[N][N];<br><span class="hljs-type">int</span> cnt[N];<span class="hljs-comment">//cnt[i]为&gt;=i的最大团点数</span><br><span class="hljs-type">int</span> group[N];<span class="hljs-comment">//最大团的点</span><br><span class="hljs-type">int</span> vis[N];<span class="hljs-comment">//记录点的位置</span><br><span class="hljs-type">int</span> res;<span class="hljs-comment">//最大团的数目</span><br><span class="hljs-function"><span class="hljs-type">bool</span> <span class="hljs-title">dfs</span><span class="hljs-params">(<span class="hljs-type">int</span> pos,<span class="hljs-type">int</span> num)</span></span>&#123;<span class="hljs-comment">//num为当前独立集中的点数</span><br>    <span class="hljs-keyword">for</span>(<span class="hljs-type">int</span> i=pos+<span class="hljs-number">1</span>;i&lt;=n;i++)&#123;<br>        <span class="hljs-keyword">if</span>(cnt[i]+num&lt;=res)<span class="hljs-comment">//剪枝，若取i但cnt[i]+已经取了的点数仍&lt;ans</span><br>            <span class="hljs-keyword">return</span> <span class="hljs-literal">false</span>;<br><br>        <span class="hljs-keyword">if</span>(G[pos][i])&#123;<span class="hljs-comment">//与当前团中元素比较，取Non-N(i)</span><br>            <span class="hljs-type">int</span> j;<br>            <span class="hljs-keyword">for</span>(j=<span class="hljs-number">0</span>;j&lt;num;j++)<br>                <span class="hljs-keyword">if</span>(!G[i][vis[j]])<br>                    <span class="hljs-keyword">break</span>;<br>            <span class="hljs-keyword">if</span>(j==num)&#123;<span class="hljs-comment">//若为空，则皆与i相邻，则此时将i加入到最大团中</span><br>                vis[num]=i;<br>                <span class="hljs-keyword">if</span>(<span class="hljs-built_in">dfs</span>(i,num+<span class="hljs-number">1</span>))<br>                    <span class="hljs-keyword">return</span> <span class="hljs-literal">true</span>;<br>            &#125;<br>        &#125;<br>    &#125;<br><br>    <span class="hljs-keyword">if</span>(num&gt;res)&#123;<span class="hljs-comment">//每添加一个点最多使最大团数+1,后面的搜索就没有意义了</span><br>        <span class="hljs-keyword">for</span>(<span class="hljs-type">int</span> i=<span class="hljs-number">0</span>;i&lt;num;i++)<span class="hljs-comment">//最大团的元素</span><br>            group[i]=vis[i];<br>        res=num;<span class="hljs-comment">//最大团中点的数目</span><br>        <span class="hljs-keyword">return</span> <span class="hljs-literal">true</span>;<br>    &#125;<br>    <span class="hljs-keyword">return</span> <span class="hljs-literal">false</span>;<br>&#125;<br><span class="hljs-function"><span class="hljs-type">void</span> <span class="hljs-title">maxClique</span><span class="hljs-params">()</span></span>&#123;<br>    res=<span class="hljs-number">-1</span>;<br>    <span class="hljs-keyword">for</span>(<span class="hljs-type">int</span> i=n;i&gt;<span class="hljs-number">0</span>;i--)&#123;<span class="hljs-comment">//枚举所有点</span><br>        vis[<span class="hljs-number">0</span>]=i;<br>        <span class="hljs-built_in">dfs</span>(i,<span class="hljs-number">1</span>);<br>        cnt[i]=res;<br>    &#125;<br>&#125;<br><span class="hljs-function"><span class="hljs-type">int</span> <span class="hljs-title">main</span><span class="hljs-params">()</span></span>&#123;<br>    <span class="hljs-type">int</span> T;<br>    <span class="hljs-built_in">scanf</span>(<span class="hljs-string">&quot;%d&quot;</span>,&amp;T);<br>    <span class="hljs-keyword">while</span>(T--)&#123;<br>        <span class="hljs-built_in">memset</span>(G,<span class="hljs-number">0</span>,<span class="hljs-built_in">sizeof</span>(G));<br><br>        <span class="hljs-built_in">scanf</span>(<span class="hljs-string">&quot;%d%d&quot;</span>,&amp;n,&amp;m);<br>        <span class="hljs-keyword">for</span>(<span class="hljs-type">int</span> i=<span class="hljs-number">0</span>;i&lt;m;i++)&#123;<br>            <span class="hljs-type">int</span> x,y;<br>            <span class="hljs-built_in">scanf</span>(<span class="hljs-string">&quot;%d%d&quot;</span>,&amp;x,&amp;y);<br>            G[x][y]=<span class="hljs-number">1</span>;<br>            G[y][x]=<span class="hljs-number">1</span>;<br>        &#125;<br><br>        <span class="hljs-comment">//建立反图</span><br>        <span class="hljs-keyword">for</span>(<span class="hljs-type">int</span> i=<span class="hljs-number">1</span>;i&lt;=n;i++)&#123;<br>            <span class="hljs-keyword">for</span>(<span class="hljs-type">int</span> j=<span class="hljs-number">1</span>;j&lt;=n;j++)&#123;<br>                <span class="hljs-keyword">if</span>(i==j)<br>                    G[i][j]=<span class="hljs-number">0</span>;<br>                <span class="hljs-keyword">else</span><br>                    G[i][j]^=<span class="hljs-number">1</span>;<br>            &#125;<br>        &#125;<br>        <span class="hljs-built_in">maxClique</span>();<br><br>        <span class="hljs-keyword">if</span>(res&lt;<span class="hljs-number">0</span>)<br>            res=<span class="hljs-number">0</span>;<br>        <span class="hljs-built_in">printf</span>(<span class="hljs-string">&quot;%d\n&quot;</span>,res);<span class="hljs-comment">//最大团的个数</span><br>        <span class="hljs-keyword">for</span>(<span class="hljs-type">int</span> i=<span class="hljs-number">0</span>;i&lt;res;i++)<span class="hljs-comment">//最大团中的顶点</span><br>            <span class="hljs-built_in">printf</span>(<span class="hljs-string">&quot;%d &quot;</span>,group[i]);<br>        <span class="hljs-built_in">printf</span>(<span class="hljs-string">&quot;\n&quot;</span>);<br>    &#125;<br>    <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;<br>&#125;<br></code></pre></td></tr></table></figure>

<p>四、刷题练手</p>
<p>1、裸题：ZOJ 1492 Maximum Clique HDU 1530</p>
<p>2、稍微麻烦点的题：HDU 3585 maximum shortest distance</p>
<p>3、一般无向图最大独立集的题目：POJ 1419 Graph Coloring</p>
<p>4、来一个染色问题：POJ 1129 Channel Allocation</p>

                
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